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Grantham PHY220 Week 2 Problem 3

A bullet is fired from a gun at a shooting range. The bullet hits the ground after 0.32 seconds. How far did it travel horizontally and vertically in this time if it was fired at a velocity of 1100 m/s?

Solution:

Assuming that the gun was fired horizontally.

Consider the horizontal component of the motion.

v_{o_x}=1100\:m/s

\Delta x=V_{0_x}t=\left(1100\:m/s\right)\left(0.32\:s\right)=352\:m

Consider the vertical component.

v_{o_y}=0\:m/s

\Delta y=V_{0_y}t-\frac{1}{2}gt^2=0-\frac{1}{2}\left(9.80\:m/s^2\right)\left(0.32\:s\right)^2=-0.50

The negative sign of  \Delta y means that the bullet went downward.

Therefore, the bullet traveled 352 m horizontally and 0.50 m vertically downward in 0.32 seconds.

Grantham PHY220 Week 2 Problem 2

An airplane is 5,000 m above an observer and 2.1 km to the west of them and 1.5 km to the north of you. Determine the angle to the plane in the x – y axis and the total distance to the plane from you. Choose the x-axis east, y axis north, and z axis up.

Solution:

The angle of the plane in the x-y axis

Solve for \theta

\theta =tan^{-1}\left(\frac{1500}{2100}\right)=35.5^{\circ}

The angle is 35.5 degrees measured from the west axis. If it is measured from the East axis, the angle would be 180-35.5=144.5 degrees.

The distance from the plane to the observer

Solve for d=\sqrt{\left(1500\:m\right)^2+\left(2100\:m\right)^2+\left(5000\:m\right)^2}=5626.72\:m